MECHANICS OF

SOLIDS CHAPTER

GIK Institute of Engineering Sciences and Technology

10 Columns

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MECHANICS OF SOLIDS

10 - 2

Columns

Stability of Structures

Euler’s Formula for Pin-Ended Beams

Extension of Euler’s Formula

Sample Problem 10.1

Design of Columns Under Centric Load

Sample Problem 10.4

Design of Columns Under an Eccentric Load

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10 - 3

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MECHANICS OF SOLIDS

10 - 4

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10 - 5

Stability of Structures

• In the design of columns, cross-sectional area is

selected such that

- allowable stress is not exceeded

allA

P

- deformation falls within specifications

specAE

PL

• After these design calculations, may discover

that the column is unstable under loading and

that it suddenly becomes sharply curved or

buckles.

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MECHANICS OF SOLIDS

10 - 6

Stability of Structures

• Consider model with two rods and torsional

spring. After a small perturbation,

moment ingdestabiliz 2

sin2

moment restoring 2

LP

LP

K

• Column is stable (tends to return to aligned

orientation) if

L

KPP

KL

P

cr4

22

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MECHANICS OF SOLIDS

10 - 7

Stability of Structures

• Assume that a load P is applied. After a

perturbation, the system settles to a new

equilibrium configuration at a finite

deflection angle.

sin4

2sin2

crP

P

K

PL

KL

P

• Noting that sin < , the assumed

configuration is only possible if P > Pcr.

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MECHANICS OF SOLIDS

10 - 8

Euler’s Formula for Pin-Ended Beams

• Consider an axially loaded beam.

After a small perturbation, the system

reaches an equilibrium configuration

such that

02

2

2

2

yEI

P

dx

yd

yEI

P

EI

M

dx

yd

• Solution with assumed configuration

can only be obtained if

2

2

2

22

2

2

rL

E

AL

ArE

A

P

L

EIPP

cr

cr

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MECHANICS OF SOLIDS

10 - 9

Euler’s Formula for Pin-Ended Beams

s ratioslendernesr

L

tresscritical srL

E

AL

ArE

A

P

A

P

L

EIPP

cr

crcr

cr

2

2

2

22

2

2

• The value of stress corresponding to

the critical load,

• Preceding analysis is limited to

centric loadings.

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MECHANICS OF SOLIDS

10 - 10

Extension of Euler’s Formula

• A column with one fixed and one free

end, will behave as the upper-half of a

pin-connected column.

• The critical loading is calculated from

Euler’s formula,

length equivalent 2

2

2

2

2

LL

rL

E

L

EIP

e

e

cr

ecr

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MECHANICS OF SOLIDS

10 - 11

Extension of Euler’s Formula

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10 - 12

Sample Problem 10.1

An aluminum column of length L and

rectangular cross-section has a fixed end at B

and supports a centric load at A. Two smooth

and rounded fixed plates restrain end A from

moving in one of the vertical planes of

symmetry but allow it to move in the other

plane.

a) Determine the ratio a/b of the two sides of

the cross-section corresponding to the most

efficient design against buckling.

b) Design the most efficient cross-section for

the column.

L = 20 in.

E = 10.1 x 106 psi

P = 5 kips

FS = 2.5

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10 - 13

Sample Problem 10.1

• Buckling in xy Plane:

12

7.0

1212

,

23121

2

a

L

r

L

ar

a

ab

ba

A

Ir

z

ze

zz

z

• Buckling in xz Plane:

12/

2

1212

,

23121

2

b

L

r

L

br

b

ab

ab

A

Ir

y

ye

yy

y

• Most efficient design:

2

7.0

12/

2

12

7.0

,,

b

a

b

L

a

L

r

L

r

L

y

ye

z

ze

35.0b

a

SOLUTION:

The most efficient design occurs when the

resistance to buckling is equal in both planes of

symmetry. This occurs when the slenderness

ratios are equal.

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10 - 14

Sample Problem 10.1

L = 20 in.

E = 10.1 x 106 psi

P = 5 kips

FS = 2.5

a/b = 0.35

• Design:

2

62

2

62

2

2

cr

cr

6.138

psi101.10

0.35

lbs 12500

6.138

psi101.10

0.35

lbs 12500

kips 5.12kips 55.2

6.138

12

in 202

12

2

bbb

brL

E

bbA

P

PFSP

bbb

L

r

L

e

cr

cr

y

e

in. 567.035.0

in. 620.1

ba

b

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10 - 15

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10 - 16

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10 - 17

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10 - 18

Design of Columns Under Centric Load

• Previous analyses assumed

stresses below the proportional

limit and initially straight,

homogeneous columns

• Experimental data demonstrate

- for large Le/r, cr follows

Euler’s formula and depends

upon E but not Y.

- for intermediate Le/r, cr

depends on both Y and E.

- for small Le/r, cr is

determined by the yield

strength Y and not E.

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10 - 19

Design of Columns Under Centric Load

Structural Steel

American Inst. of Steel Construction

• For Le/r > Cc

92.1

/2

2

FS

FSrL

E crall

e

cr

• For Le/r < Cc

3

2

2

/

8

1/

8

3

3

5

2

/1

c

e

c

e

crall

c

eYcr

C

rL

C

rLFS

FSC

rL

• At Le/r = Cc

YcYcr

EC

22

21 2

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10 - 20

Design of Columns Under Centric Load

Aluminum

Aluminum Association, Inc.

• Alloy 6061-T6

Le/r < 66:

MPa /868.0139

ksi /126.02.20

rL

rL

e

eall

Le/r > 66:

2

3

2/

MPa 10513

/

ksi 51000

rLrL ee

all

• Alloy 2014-T6

Le/r < 55:

MPa /585.1212

ksi /23.07.30

rL

rL

e

eall

Le/r > 55:

2

3

2/

MPa 10273

/

ksi 54000

rLrL ee

all

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10 - 21

Sample Problem 10.4

Using the aluminum alloy2014-T6,

determine the smallest diameter rod

which can be used to support the centric

load P = 60 kN if a) L = 750 mm,

b) L = 300 mm

SOLUTION:

• With the diameter unknown, the

slenderness ration can not be evaluated.

Must make an assumption on which

slenderness ratio regime to utilize.

• Calculate required diameter for

assumed slenderness ratio regime.

• Evaluate slenderness ratio and verify

initial assumption. Repeat if necessary.

GIK Institute of Engineering Sciences and Technology

MECHANICS OF SOLIDS

10 - 22

Sample Problem 10.4

2

4

gyration of radius

radiuscylinder

2

4 c

c

c

A

I

r

c

• For L = 750 mm, assume L/r > 55

• Determine cylinder radius:

mm44.18

c/2

m 0.750

MPa 103721060

rL

MPa 10372

2

3

2

3

2

3

cc

N

A

Pall

• Check slenderness ratio assumption:

553.81

mm 18.44

mm750

2/

c

L

r

L

assumption was correct

mm 9.362 cd

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MECHANICS OF SOLIDS

10 - 23

Sample Problem 10.4

• For L = 300 mm, assume L/r < 55

• Determine cylinder radius:

mm00.12

Pa102/

m 3.0585.1212

1060

MPa 585.1212

62

3

c

cc

N

r

L

A

Pall

• Check slenderness ratio assumption:

5550

mm 12.00

mm 003

2/

c

L

r

L

assumption was correct

mm 0.242 cd

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MECHANICS OF SOLIDS

10 - 24

Design of Columns Under an Eccentric Load

• Allowable stress method:

allI

Mc

A

P

• Interaction method:

1

bendingallcentricall

IMcAP

• An eccentric load P can be replaced by a

centric load P and a couple M = Pe.

• Normal stresses can be found from

superposing the stresses due to the centric

load and couple,

I

Mc

A

P

bendingcentric

max